TS EAMCET 4 May 2018 Shift 2 Solved Paper

Consider the series,

3

10

+

3.7

10.15

+

3.7.9

10.15.20

+..... =

3

5×2!

+

3.7

52.3!

+

3.7.9

53.4!

+.....
Consider the expression,
(1+x)n=1+nx+

n(n−1)

2!

x2 +

n(n−1)(n−2)

3!

x3+.......
On comparison, we get

n(n−1)

2!

x2=

3

5×2!

And,

n(n−1)(n−2)

3!

x3=

3.7

52×3!

n(n−1)x2=

3

5

........(I)
And,
n(n−1)(n−2)x3=

21

25

.......(II)
This implies,
(n−2)x(

3

5

)=

21

25

x=

7

5(n−2)

Substitute the value of x in equation (1).
n(n−1)

49

25(n−2)2

=

3

5

34n2+11m−60=0
This gives,
n=−

3

2

or n=

20

17

So,
x=

7

5

1

(− 3 2 −2)

=−

2

5

The sum of the series is given by,
(1−

2

5

)

−3

2

−1−

3

5

=(

5

3

)

3

2

−

8

5

=

5√5

3√3

−

8

5